回复：求一段简短的快排代码

5L 少了一个end; {if}

高三党好无奈啊...练习的时间太少了...
明天再来看...

#include <iostream>
using namespace std;
struct node { int left, right; };
int main (){
int top = 0, data[10];
node stack[100];
for (int i1 = 0; i1 < 10; i1 ++)
cin >> data[i1];
stack[++ top].left = 0, stack[top].right = 9;
int left, right, flag;
node tmpPoi;
while (top > 0) {
tmpPoi = stack[top --];
left = tmpPoi.left;
right = tmpPoi.right;
flag = data[left];
while (left < right) {
while (left < right && data[right] < flag) right --;
if (left < right) data[left] = data[right], left ++;
while (left < right && data[left] > flag) left ++;
if (left < right) data[right] = data[left], right --;
}
data[left] = flag;
left --, right ++;
if (left > tmpPoi.left)
stack[++ top].left = tmpPoi.left, stack[top].right = left;
if (right < tmpPoi.right)
stack[++ top].left = right, stack[top].right = tmpPoi.right;
}
for (int i1 = 0; i1 < 10; i1 ++)
cout << data[i1] << ' ';
return 0;
}
以前写的，当然注意一下data数组的大小⋯⋯
还有stack栈的大小⋯⋯
改一下不是很难⋯⋯
非递归⋯⋯但是没加优化⋯⋯


话说其实是老师让我们自己出初赛模拟题⋯⋯
所以就没考虑扩展⋯⋯

FPC安装文件夹的DEMO里明明就有的个说= =